If d.c. unregulated input is 20 V, V0 = 10 V, load current is 0–20 mA, design the regulator, assume for the Zener, Iz(min) = 10 mA, Iz(max) = 100 mA.

Question

If d.c. unregulated input is 20 V, [latex]V_{0}[/latex] = 10 V, load current is 0–20 mA, design the regulator, assume for the Zener, [latex]I_{z(min)}[/latex] = 10 mA, [latex]I_{z(max)}[/latex] = 100 mA.

Accepted Answer

Given input voltage, [latex]V_{i}[/latex] = 20 V
Output voltage [latex]V_{0}[/latex] = 10 V
Load current varies from 0 to 20 mA

[latex]I_{z(min)}[/latex] = 10 mA, [latex]I_{z(max)}[/latex] = 100 mA

Here, [latex]V_{z} = V_{0}[/latex] = 10 V (constant)
Applying KVL to a closed-loop circuit,
20 = IR + 10
or IR = 10

Therefore, R = [latex]\frac{10}{I}[/latex] Ω, where I is the loop current in amperes

(i) Let [latex]I_{z} = I_{z(min)}[/latex] and [latex]I_{L}[/latex] = 0
The total current I = [latex]I_{L} + I_{z}[/latex] = 10 mA
Therefore, R = [latex]\frac{10\,V}{10\,mA}[/latex] = 1000 Ω

(ii) For [latex]I_{z} = I_{z(max)}[/latex] = 100 mA and [latex]I_{L}[/latex] = 20 mA
I = [latex]I_{L} + I_{z}[/latex] = 20 + 100 = 120 mA
Therefore R = [latex]\frac{10\,V}{120\,mA}[/latex] = 83.33 Ω
(iii) The range of R varies from 83.33 Ω to 1000 Ω.

Question 18.33: If d.c. unregulated input is 20 V, V0 = 10 V, load current i...

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